Answer:
The 4th term of the expanded binomial is [tex]-4320x^3y^3[/tex]
Step-by-step explanation:
Considering:
[tex]$ (x+y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k}y^k$[/tex]
[tex]$ (2x-3y)^6 = \sum_{k=0}^{6} \binom{6}{k} (2x)^{6-k}(-3y)^k$[/tex]
Now, you gotta calculate for every value of [tex]k[/tex]
[tex]$ (2x-3y)^6 = \binom{6}{0} (2x)^{6-0}(-3y)^0 + \binom{6}{1} (2x)^{6-1}(-3y)^1 + \binom{6}{2} (2x)^{6-2}(-3y)^2 + \\ $[/tex]
[tex]$\binom{6}{3} (2x)^{6-3}(-3y)^3 + \binom{6}{4} (2x)^{6-4}(-3y)^4 + \binom{6}{5} (2x)^{6-5}(-3y)^5 + \binom{6}{6} (2x)^{6-6}(-3y)^6 $[/tex]
I will not write every product, but just solve following the steps:
For [tex]k=0[/tex]
[tex]$\binom{6}{0} (2x)^{6-0}(-3y)^0$[/tex]
[tex]$\frac{6!}{(6-0)!(0!)} (2x)^{6-0}(-3y)^0$[/tex]
[tex]$ \frac{6!}{6!} \left(2x\right)^{6-0}\cdot 1$[/tex]
[tex]$1\cdot \:1\cdot \left(2x\right)^{6-0}$[/tex]
[tex]$2^6x^6$[/tex]
[tex]64x^6[/tex]
[tex](2x-3y)^6=64x^6-576x^5y+2160x^4y^2-4320x^3y^3+4860x^2y^4-2916xy^5+729y^6[/tex]
